Binomial expansion: $$(a+bx)^n$$​​

In a nutshell

You can use the formula for the expansion of $$(1+x)^n$$​ to compute the expansion of $$(a+bx)^n$$ for any constants $$a$$​ and $$b$$​. In the case that ​$$n$$ is not a non-negative integer, this expansion will only be valid when $$|x| < \left| \dfrac{a}{b}\right|$$​.

Using the expansion of $$(1+x)^n$$​​

The expansion formula tells you that:

​$$(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dots + \binom{n}{r}x^r + \dots$$​​

This holds in general whenever $$n$$​ is a non-negative integer, as in these cases only finitely many binomial coefficients are nonzero, whereas when $$n$$​ is not a non-negative integer this produces an infinite series, and only holds whenever $$|x| < 1$$​.

You can use this to derive the binomial expansion for $$(a+bx)^n$$​ using this formula. If $$a=0$$​, then $$(a+bx)^n = b^nx^n$$​, but otherwise:

​$$\begin{aligned}(a + bx)^n &= \left(a\left(1+\dfrac{b}{a}x\right)\right)^n\\&= a^n\left(1 + \dfrac{b}{a}x\right)^n \\&= a^n\left[1 + n\left(\dfrac{b}{a}x\right) + \dfrac{n(n-1)}{2!}\left(\dfrac{b}{a}x\right)^2 + \dots + \dfrac{n(n-1)(n-2)...(n-r+1)}{r!}\left(\dfrac{b}{a}x\right)^n + \dots\right]\end{aligned}$$​​

This holds in general whenever $$n \in \mathbb{N}$$​, and if $$n$$​ is any other real value this only holds when $$\left|\dfrac{b}{a}x\right| < 1$$​, i.e. when $$|x| < \left|\dfrac{a}{b}\right|$$​.

Example 1

Find the first $$4$$​ terms in the expansion of $$\dfrac{1}{\sqrt{12 - 3x}}$$​, and determine the range of values of $$x$$ for which this expansion holds.

Rearrange $$\dfrac{1}{\sqrt{12-3x}}$$:

$$\begin{aligned}\dfrac{1}{\sqrt{12 - 3x}} &= (\sqrt{12-3x})^{-1}\\&= \left(12-3x\right)^{-\dfrac12}\\&= (12)^{-\dfrac12}\left(1 - \dfrac14x\right)^{-\dfrac12}\end{aligned}$$​​

​

Compute the first $$4$$ terms using the expansion of $$\left(1 - \dfrac14x\right)^{-\dfrac12}$$:

​$$\begin{aligned}\dfrac{1}{\sqrt{12-3x}} &= (12)^{-\dfrac12}\left(1-\dfrac14x\right)^{-\dfrac12}\\&=(12)^{-\dfrac12}\left[1 +\left(-\dfrac12\right)\left(- \dfrac14x\right) + \dfrac{\left(-\dfrac12\right) \left(-\dfrac32\right)}{2!}\left(-\dfrac14x\right)^2 + \dfrac{\left(-\dfrac12\right)\left(-\dfrac32\right)\left(-\dfrac52\right)}{3!}\left(-\dfrac14x\right)^3 + \dots\right]\\&= (12)^{-\dfrac12}\left[1 + \dfrac{(-1)^2}{2\times 4}x + \dfrac{(-1)^4\times3}{2^3 \times 4^2}x^2 + \dfrac{(-1)^6 \times 3 \times 5}{2^3 \times 6 \times 4^3}x^3 + \dots\right]\\&=\dfrac{\sqrt{3}}{6}\left[1 + \dfrac{1}{8}x + \dfrac{3}{128}x^2 + \dfrac{15}{3072}x^3 + \dots\right]\\&= \dfrac{\sqrt{3}}{6} + \dfrac{\sqrt{3}}{48}x + \dfrac{\sqrt{3}}{256}x^2 + \dfrac{5\sqrt{3}}{6144}x^3 + \dots\end{aligned}$$​​

This expansion is valid when $$\left|-\dfrac14x\right| < 1$$, i.e. when $$|x| < 4$$.

Therefore, the first four terms of the expansion of $$\dfrac{1}{\sqrt{12-3x}}$$ are $$\underline{\dfrac{\sqrt{3}}{6} + \dfrac{\sqrt{3}}{48}x + \dfrac{\sqrt{3}}{256}x^2 + \dfrac{5\sqrt{3}}{6144}x^3 + \dots}$$, and the range of values of $$x$$ for which this expansion holds is $$\underline{|x|< 4}$$.